 Solution

Let $m$ be the car's mass and let $v(t)$ be its forward speed. Initially the driven contact surfaces slip backward relative to the ground because $v<r\Omega$. Thus kinetic friction on the car acts forward. On the horizontal surface the total normal force is $mg$, so Newton's second law gives $m\dot{v}=\mu mg$. While slipping persists,The contact slip speed is $r\Omega -v(t)$. Its first zero is the onset of rolling without slipping, henceThe engine maintains the prescribed wheel speed, so wheel spin-down is not part of this calculation.

On a slope of angle $\alpha >0$, directed uphill, the normal force is $mg\cos \alpha$ and gravity contributes $-mg\sin \alpha$ along the slope. Therefore the uphill acceleration while slipping is $g(\mu \cos \alpha -\sin \alpha )$. If it is positive, the onset of rolling without slipping occurs atThe inequality follows because $\mu \cos \alpha -\sin \alpha <\mu$. If $\mu \cos \alpha \le \sin \alpha$, the car does not accelerate uphill to $r\Omega$, so there is no finite catch-up time in this sliding-friction model. The finite-time conclusion presumes positive traction sufficient to climb the slope.